Given the adjoint matrix A ^ * = diag (1,1,1,8), and ABA ^ - 1 = Ba ^ - 1 + 3E, find B

Given the adjoint matrix A ^ * = diag (1,1,1,8), and ABA ^ - 1 = Ba ^ - 1 + 3E, find B

From the known ABA ^ - 1 = Ba ^ - 1 + 3E equation, we get that | a | B = a * B + 3 | a | e because | a | = 8 = | a | ^ 3, so | a | = 2, so 2B = a * B + 6e, so (2e-a *) B = 6e, so B = 6 (2e-a *) ^ - 1 = 6diag (1,1, - 1 / 6) = diag (6,6

Given the adjoint matrix A * = (1000) (0100) (1010) (1 - 308) and ABA ^ (- 1) = Ba ^ (- 1) + 3E of matrix A, we can find B
The adjoint matrix A * = (100) of known matrix A
(0 1 0 0)
(1 0 1 0)
(1 -3 0 8)
And ABA ^ (- 1) = Ba ^ (- 1) + 3E
Seeking B

From aba ^ (- 1) = Ba ^ (- 1) + 3e
Multiply a on both sides right to get AB = B + 3a
Multiply a * left on both sides to get | a | B = a * B + 3 | a | E
8 = |A*|=|A|^3
So | a | = 2
So 2B = a * B + 6e
That is, (2e-a *) B = 6e
So B = 6 (2e-a *) ^ - 1
(2E-A*,E)
1 0 0 0 1 0 0 0
0 1 0 0 0 1 0 0
-1 0 1 0 0 0 1 0
-1 3 0 -6 0 0 0 1
r3+r1,r4+r1
1 0 0 0 1 0 0 0
0 1 0 0 0 1 0 0
0 0 1 0 1 0 1 0
0 3 0 -6 1 0 0 1
r4-3r2
1 0 0 0 1 0 0 0
0 1 0 0 0 1 0 0
0 0 1 0 1 0 1 0
0 0 0 -6 1 -3 0 1
r4*(-1/6)
1 0 0 0 1 0 0 0
0 1 0 0 0 1 0 0
0 0 1 0 1 0 1 0
0 0 0 1 -1/6 1/2 0 -1/6
So B = 6 (2e-a *) ^ - 1=
6 0 0 0
0 6 0 0
6 0 6 0
-1 3 0 -1

On the determinant calculation of diagonal matrix in linear algebra
The determinant value on the main diagonal is the product of the numbers on the main diagonal. But the sub diagonal needs to be preceded by a (- 1) n (n-1) / 2? I'll calculate the exponent that (n + 1) * (n-1) is - 1 by myself. What's the result

The determinant can be expanded by row: the first row is reduced step by step, the first (- 1) ^ (n + 1), the second (- 1) ^ n, the third (- 1) ^ (n-1) The last time (- 1) ^ 3 appears. Therefore, the sign of the coefficient is (- 1) ^ [(n + 1) + N + +3] In this paper, we discuss the relationship between the two factors

How to prove that diag [a, B, C] is similar to diag [C, B, a]?

Order p=
0 0 1
0 1 0
1 0 0
Then p ^ - 1diag [a, B, C] P = diag [C, B, a]